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Mseemab
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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h 25 Jun 2019, 05:04
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B
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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In how many ways can we pick two books without replacement such that both books have different colors?

A. 28
B. 63
C. 25
D. 52
E. 60

Experts please explain, what can be the possible ways to approach such questions?
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B
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MahmoudFawzy
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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h Updated on: 25 Jun 2019, 06:32
Originally posted by MahmoudFawzy on 25 Jun 2019, 06:16.
Last edited by MahmoudFawzy on 25 Jun 2019, 06:32, edited 1 time in total.
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For books to have different colors, there are three ways:
red+green possibilities = 3*5 = 15
green+blue possibilities = 5*6 = 30
red+blue possibilities = 3*6 = 18

so the total ways = 15 + 18 + 30 = 63
IMO B

A reverse method can be done by calculating (the total ways of choosing two books) - (the ways of choosing similar books):

14C2 - (3C2 + 5C2 + 6C2) = 91 - (3+10+15) = 91 - 28 = 63
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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h 25 Jun 2019, 06:31
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Total Number of ways of picking two books = 14C2 = 91

Number of ways both books are red = 3C2 = 3

Number of ways both books are green = 5C2 = 10

Number of ways both books are blue = 6C2 = 15

Total number ways the two books are not the same color = 91-(3+10+15) = 63

Answer is (B)

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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h 25 Jun 2019, 08:38
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To get differrent colours, let see the differrent possibiities:
Red and green = 3C1 * 5C1 = 15
Red and blue = 3C1 * 6C1 = 18
Green and blue = 5C1 * 6C1 = 30
No. Of ways of picking two differrent colours = 15+18+30 = 63.
I hope this is helpful

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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h 01 Jul 2019, 18:06
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Mseemab
There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In how many ways can we pick two books without replacement such that both books have different colors?

A. 28
B. 63
C. 25
D. 52
E. 60

Experts please explain, what can be the possible ways to approach such questions?


The total number of ways to select two books from 14 is 14C2 = 14! / (12! x 2!) = (14 x 13)/2 = 7 x 13 = 91.

The number of ways to select 2 red books is 3C2 = 3.

The number of ways to select 2 green books is 5C2 = 5! / (3! x 2!) = (5 x 4)/2 = 10.

The number of ways to select 2 blue books is 6C2 = 6! / (4! x 2!) = (6 x 5)/2 = 15.

So the number of ways to select books of different colors is:

91 - 10 - 15 - 3 = 91 - 28 = 63

Answer: B
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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h 19 Jul 2019, 14:07
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Total ways to select:
1 red + 1 any other = 3 x 11 = 33
1 green + 1 any other = 5 x 9 = 45
1 blue + 1 any other = 6 x 8 = 48

Total ways = 126

Divide by 2 (since blue/green and green/blue is the same) , therefore 63

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There are 3 Red Books, 5 Green Books and 6 Blue Books on a shelf. In h 12 May 2024, 01:25
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